I have the following problem that I am stuck on:

Consider the element $a=\sqrt[3]{7}$ of $\mathbf{R}$. Show that this element is algebraic over $\mathbf{Q}$ and find its minimal polynomial. Also, find the degree of the extension $[\mathbf{Q}\left(\sqrt[3]{7}\right):\mathbf{Q}]$ and find a basis of $\mathbf{Q}\left(\sqrt[3]{7}\right)$ over $\mathbf{Q}$.

My thoughts so far: I think that the minimal polynomial is $f(x)=x^{3}-7$. Is this correct? If so, how to I prove this is the minimal polynomial? Also, I think that the basis should be $\{1,\sqrt[3]{7},\sqrt[3]{49}\}$ based on some other examples I have seen. Is this correct? If so, could someone explain why this is the correct basis? If not, what is the basis?

Thanks in advance for any help!


Of course $f(x) = x^3-7$ has $\sqrt[3]{7}$ as a root, so you just need to show that $f$ is irreducible to show that it is indeed the minimal polynomial of $\sqrt[3]{7}$. What irreducibility tests do you know?

Also, your basis is correct. Suppose we have an algebraic extension $\mathbb{Q}(\alpha)/ \mathbb{Q}$. If the degree of the extension is $n$, then $\{\alpha^k\}_{k=0}^{n-1}$ serves as a basis.

Proof: Suppose we could write $\displaystyle \sum_{k=0}^{n-1} c_k \alpha^k = 0$ for $c_k \in \mathbb{Q}$ not all zero. Then $\displaystyle \sum_{k=0}^{n-1} c_kx^k$ is a polynomial for which $\alpha$ is a root: a contradiction because the minimal polynomial for $\alpha$ must have degree $n = [ \mathbb{Q}(\alpha) : \mathbb{Q}]$. Furthermore, that we must have $c_k = 0$ for all $0 \leq k \leq n-1$ demonstrates the linear independence of the set $\{\alpha^k\}_{k=0}^{n-1}$. The cardinality of that linearly independent set is equal to the degree of the extension, so we can rest assured that it is indeed a basis for the extension**.

**Just to drive home fundamentally important ideas: recall that the degree of an extension $K/F$ is defined to be the dimension of $K$ when $K$ is viewed as a vector space over $F$. The dimension of a vector space over a field is defined to be the cardinality of a given basis for the vector space. And it is a theorem that all possible bases of a vector space have equal cardinality, and that any linearly independent set of vectors with that cardinality is necessarily a basis.

  • $\begingroup$ That helps! Thank you! $\endgroup$ – Sir_Math_Cat May 2 '17 at 3:12
  • $\begingroup$ Glad I could help @A.S.Hopkins. I fixed some minor typos, wherein I had $k=1$ instead of $k=0$ as the bounds in the summation, and I had written $\alpha_k$ instead of $\alpha^k$ in a couple spots. $\endgroup$ – Kaj Hansen May 2 '17 at 21:09

Since $a$ is clearly a root of $f$, it's enough to show that $f$ is irreducible over $\mathbb{Q}$. Since $f$ is cubic, it suffices to show that $f$ has no roots in $\mathbb{Q}$.

  • $\begingroup$ I figured that part out already. Is this all we have to do to show that it is the minimal polynomial? Like nothing else to show uniqueness or anything? $\endgroup$ – Sir_Math_Cat May 2 '17 at 3:06
  • $\begingroup$ The uniqueness of the minimal polynomial follows from the definition. $\endgroup$ – Ethan Alwaise May 2 '17 at 3:09
  • $\begingroup$ Okay, that makes sense. How would you determine the basis of the extension? $\endgroup$ – Sir_Math_Cat May 2 '17 at 3:10

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